1. **Problem Statement:** Find the derivative $\frac{dy}{dx}$ for the functions using the definition of derivative and verify using derivative formulas. 2. **Definition of Derivative:** The derivative of $y=f(x)$ at $x$ is given by $$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This represents the slope of the tangent line to the curve at point $x$. 3. **Part (a):** $y = \sqrt{9 - 4x}$ - Using the definition: $$f(x) = \sqrt{9 - 4x}$$ $$f(x+h) = \sqrt{9 - 4(x+h)} = \sqrt{9 - 4x - 4h}$$ Calculate difference quotient: $$\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{9 - 4x - 4h} - \sqrt{9 - 4x}}{h}$$ Multiply numerator and denominator by the conjugate: $$= \frac{(9 - 4x - 4h) - (9 - 4x)}{h(\sqrt{9 - 4x - 4h} + \sqrt{9 - 4x})} = \frac{-4h}{h(\sqrt{9 - 4x - 4h} + \sqrt{9 - 4x})}$$ Simplify: $$= \frac{-4}{\sqrt{9 - 4x - 4h} + \sqrt{9 - 4x}}$$ Take limit as $h \to 0$: $$\frac{dy}{dx} = \lim_{h \to 0} \frac{-4}{\sqrt{9 - 4x - 4h} + \sqrt{9 - 4x}} = \frac{-4}{2\sqrt{9 - 4x}} = \frac{-2}{\sqrt{9 - 4x}}$$ - Using derivative formulas: $$y = (9 - 4x)^{1/2}$$ $$\frac{dy}{dx} = \frac{1}{2}(9 - 4x)^{-1/2} \cdot (-4) = \frac{-2}{\sqrt{9 - 4x}}$$ 4. **Part (b):** $y = \frac{x}{x+1}$ - Using the definition: $$f(x) = \frac{x}{x+1}$$ $$f(x+h) = \frac{x+h}{x+h+1}$$ Difference quotient: $$\frac{f(x+h) - f(x)}{h} = \frac{\frac{x+h}{x+h+1} - \frac{x}{x+1}}{h} = \frac{(x+h)(x+1) - x(x+h+1)}{h(x+h+1)(x+1)}$$ Expand numerator: $$(x+h)(x+1) = x^2 + x + hx + h$$ $$x(x+h+1) = x^2 + xh + x$$ Subtract: $$x^2 + x + hx + h - (x^2 + xh + x) = h$$ So difference quotient: $$\frac{h}{h(x+h+1)(x+1)} = \frac{1}{(x+h+1)(x+1)}$$ Take limit as $h \to 0$: $$\frac{dy}{dx} = \lim_{h \to 0} \frac{1}{(x+h+1)(x+1)} = \frac{1}{(x+1)^2}$$ - Using quotient rule: $$y = \frac{u}{v}, u = x, v = x+1$$ $$\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{x+1 - x}{(x+1)^2} = \frac{1}{(x+1)^2}$$ **Final answers:** (a) $$\frac{dy}{dx} = \frac{-2}{\sqrt{9 - 4x}}$$ (b) $$\frac{dy}{dx} = \frac{1}{(x+1)^2}$$